[QUOTE=jyb;449303]Y0 and Y1 are the coefficients of the rational polynomial, which is assumed to be of degree 1. So that polynomial is just Y1*x + Y0. If you specify the value of m directly, as in your case, then Y1 is assumed to be 1 and Y0 is assumed be be -m (or -1 and m, respectively).

Giving the coefficients in this way may be more convenient than specifying m because it can allow you to avoid any modular inversions. For your number above it wouldn't make any difference, but it comes up when the number has more of a homogeneous form.

For example, to find a sextic polynomial for something like a^131 - b^131, you would multiply by ab to get b*a^132 - a*b^132, which is bx^6 - ay^6, where x = a^22 and y = b^22. Dividing both sides by y^6 gives you a polynomial in the single variable (x/y). So now your 'm' value is a^22/b^22. If you want to specify that directly, you need to do a modular inversion of b mod n (or of b^22 mod n if that's easier for some reason). But your rational polynomial, instead of being x - a^22/b^22, can instead be (b^22)*x - a^22. So you can just set Y1 = b^22 and Y0 = -a^22 and leave out m.[/QUOTE]

Excellent! I've seen the m = x/y description before, but this clears my understanding of Y0 and Y1 for both my Riesel-number case and the homogeneous-Cunningham case.
Thank you very much.

It's probably also worth pointing out that the modular inverse of b mod n would be a number about the size of n, whilst b^22 and a^22 are both about the size of the sixth root of n. Doing the modular inversion explicitly would give you a rational polynomial with coefficients too large for NFS to work in a reasonable length of time.

125!+1
 

[URL="http://www.mersenneforum.org/showpost.php?p=407860&postcount=462"]A polynomial has been requested[/URL] for the remaining [URL="http://www.factordb.com/index.php?id=1100000000009752382"]C184[/URL] cofactor
of [URL="http://www.factordb.com/index.php?id=1000000000002850149"]125!+1[/URL] however it appears to me that the number is not yet
factored (a bit more than one year after the polynomial was requested).
Wombatman answered with pretty decent polynomials, I'm just wondering
whether the idea of queuing this number for sieving by NFS@home
has been abandoned.. Any thoughts?

14e candidate
 

C208_127_84 - survived 6k curves @B1=11e7 plus 1.5k curves @B1=3e8. A bit light for ECM by the 2/9 rule but overkill by the total time of ECM < 1/3 * NFS time criteria, as this number sieves very quickly.

[code]
n: 5899230812152449499439060523411070615862912267516217887300205504079653042931928206512510869381602808260604401867058923090523452877179689722846978468337799634433298635474866355567751092498919104869955765192127
# 127^84+84^127, difficulty: 244.38, anorm: 1.83e+037, rnorm: -3.72e+046
# scaled difficulty: 245.93, suggest sieving rational side
# size = 6.096e-012, alpha = 0.000, combined = 4.413e-013, rroots = 0
type: snfs
size: 244
skew: 2.0927
c6: 1
c0: 84
Y1: -25695969452033992329379379343259582070784
Y0: 283956682347124706942551243009
rlim: 134000000
alim: 134000000
lpbr: 31
lpba: 31
mfbr: 62
mfba: 62
rlambda: 2.7
alambda: 2.7
[/code]

[QUOTE=YuL;449351][URL="http://www.mersenneforum.org/showpost.php?p=407860&postcount=462"]A polynomial has been requested[/URL] for the remaining [URL="http://www.factordb.com/index.php?id=1100000000009752382"]C184[/URL] cofactor
of [URL="http://www.factordb.com/index.php?id=1000000000002850149"]125!+1[/URL] however it appears to me that the number is not yet
factored (a bit more than one year after the polynomial was requested).
Wombatman answered with pretty decent polynomials, I'm just wondering
whether the idea of queuing this number for sieving by NFS@home
has been abandoned.. Any thoughts?[/QUOTE]

I'm not aware that anybody has explicitly decided to abandon it, I just think that people forgot about it.

There have been other GNFS jobs with numbers around this size with 15e, so it should be fine, though of course the 15e queue isn't hurting for candidates at the moment. Still, if you can do some test sieving to determine good parameters for the job, it can be queued.

14e running dry
 

Anybody have any suitable candidates ready for 14e? I will have one soon but it still needs some more ECM.

I have two more OPN quintics from the [URL=http://www.lirmm.fr/~ochem/opn/t490.txt]t490[/URL] file.

208097431^29-1 (C229) SNFS-250
790579404481^19-1 (C192) SNFS-238

The second one may seem a little small but it doesn't sieve well because of the large coefficient. So it may still be a good 14e candidate.

[QUOTE=RichD;449587]I have two more OPN quintics from the [URL=http://www.lirmm.fr/~ochem/opn/t490.txt]t490[/URL] file.

208097431^29-1 (C229) SNFS-250
790579404481^19-1 (C192) SNFS-238

The second one may seem a little small but it doesn't sieve well because of the large coefficient. So it may still be a good 14e candidate.[/QUOTE]

Did you do any test sieving with these? They both seem to sieve very badly. Difficulty 250 is normally reasonable for 14e, but the first one appears much better with 15e. And how did you calculate a difficulty of 238 for the second one? I get 226. But even with that it doesn't sieve very well (as you already mentioned). It can be a 14e job, though.

Edit: The second one also sieves better with special Q on the algebraic side.

Queueing them both...

Queueing another HCN, 9+5_295.

I've been experimenting with quartics to see what the right difficulties are for the sievers. This one is an SNFS-225, and it already appears to hit a crossover point: it's more efficient with the 15e siever. This has had ECM of about 2/3 of a t55.

[QUOTE=jyb;449632]Did you do any test sieving with these? They both seem to sieve very badly. Difficulty 250 is normally reasonable for 14e, but the first one appears much better with 15e. And how did you calculate a difficulty of 238 for the second one? I get 226. But even with that it doesn't sieve very well (as you already mentioned). It can be a 14e job, though.

Edit: The second one also sieves better with special Q on the algebraic side.

Queueing them both...[/QUOTE]

I did test sieving to see if it is better to create a degree 4, 5 or 6 polynomial (if applicable). but only on the rational side. That is why I mentioned "both as quintic" meaning the degree 5 indicated the best results.

The second one was elevated to degree 5 (P^20-P) thus calculating at SNFS-238 or there about. After thinking about it, it doesn't surprise me it sieves better on the algebraic side.

Through some limited testing of my own, and at higher difficulties (i.e. 15e range), I found that large coefficients may sieve better on the algebraic at the lower special-Q values. Given that more relations are needed through greater Q values, the algebraic side seems to peter out faster. If only one side is picked to sieve on, it may be advantageous to use the rational side for the total job for larger numbers.

Which bring me to the next question. Sometimes it is optimal to sieve on both sides. Meaning, let's say, 1/3 on one side and 2/3 on the other side to collect the relations needed. Is this something NFS@Home is capable of or is it too much trouble?

[QUOTE=RichD;449644]I did test sieving to see if it is better to create a degree 4, 5 or 6 polynomial (if applicable). but only on the rational side. That is why I mentioned "both as quintic" meaning the degree 5 indicated the best results.

The second one was elevated to degree 5 (P^20-P) thus calculating at SNFS-238 or there about. After thinking about it, it doesn't surprise me it sieves better on the algebraic side.

Through some limited testing of my own, and at higher difficulties (i.e. 15e range), I found that large coefficients may sieve better on the algebraic at the lower special-Q values. Given that more relations are needed through greater Q values, the algebraic side seems to peter out faster. If only one side is picked to sieve on, it may be advantageous to use the rational side for the total job for larger numbers.
[/QUOTE]

Well this is strange. Because I stupidly failed to notice that you had said these should both be quintics, I actually test-sieved them both as sextics (which is why I got the lower difficulty for the second one). I just tried them as quintics, and they are both [I]much[/I] worse, whether on the rational or algebraic side.

So now I'm confused. You got better results with quintics, but I've gotten much better results with sextics. What gives?

[QUOTE=RichD;449644]
Which bring me to the next question. Sometimes it is optimal to sieve on both sides. Meaning, let's say, 1/3 on one side and 2/3 on the other side to collect the relations needed. Is this something NFS@Home is capable of or is it too much trouble?[/QUOTE]

I don't have access to enough of the inner workings to have a definitive answer to this. All I know is that the method I've been given to submit jobs allows for a single polynomial with parameters, including on which side to sieve special-q. So as far as my abilities go, no it is not possible to do what you're suggesting.